(a) how high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?
(b) if, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?
(c) what is the average force of friction if the hill has a slope of 2.5º above the horizontal?

Explanation:

a ) Let the height achieved be h .

We shall apply law  of conservation of mechanical energy.

1 /2 mv² = mgh

h = v² / 2g

v = 110 km/h

= 30.55 m /s

h =

h = 47.61 m

b )

Kinetic energy of car in the beginning

= 1/2 x 750 x (30.55)²

= 349988.43 J

Potential energy at 22 m height

= 750 x 9.8 x 22

= 161700 J

Energy lost due to frictional force

= 349988.43 - 161700

= 188288.43 J

c )

Distance covered along the slope

d = 22 / sin2.5

= 22 / 0.043619

d = 504.36 m

If F be average frictional force

work done by friction

F x d

= F x 504.36

so

F x 504.36 = 188288.43

F = 188288.43 / 504.36

= 373.32 N